A wire of length 1 m and radius `10^-3`m is carrying a heavy current and is assumed to radiate as a black body. At equilibrium, its temperature is 900 K while that of surrounding is 300 K. The resistivity of the material of the wire at 300 K is `pi^@xx10^-8` ohm m and its temperature coefficient of resistance is `7.8xx10^(-3)//C` (stefan's constant`sigma=5.68xx10^-8W//m^(2) K^(2)`.
Heat radiated per second by the wire is nearly

Options

A. 23 W
B. 230 W
C. 2300 W
D. 23000 W

Explanation

To solve the problem of finding the heat radiated per second by the wire, we will use the Stefan-Boltzmann law, which states that the power radiated by a black body per unit area is proportional to the fourth power of its temperature. The formula for the power radiated by a black body is given by:

\[ P = \sigma A (T^4 - T_{surrounding}^4) \]

Where:
- \( P \) is the power radiated (in watts),
- \( \sigma \) is the Stefan-Boltzmann constant (\( 5.68 \times 10^{-8} \, \text{W/m}^2 \, \text{K}^4 \)),
- \( A \) is the surface area of the wire (in m²),
- \( T \) is the temperature of the wire (in K),
- \( T_{surrounding} \) is the temperature of the surroundings (in K).

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of the wire, \( L = 1 \, \text{m} \)
- Radius of the wire, \( r = 10^{-3} \, \text{m} \)
- Temperature of the wire, \( T = 900 \, \text{K} \)
- Temperature of the surroundings, \( T_{surrounding} = 300 \, \text{K} \)
- Stefan-Boltzmann constant, \( \sigma = 5.68 \times 10^{-8} \, \text{W/m}^2 \, \text{K}^4 \)

2. **Calculate the surface area of the wire:**
The surface area \( A \) of a cylindrical wire can be calculated using the formula:
\[ A = 2 \pi r L \]
Substituting the values:
\[ A = 2 \pi (10^{-3}) (1) = 2 \pi \times 10^{-3} \, \text{m}^2 \]

3. **Calculate \( T^4 \) and \( T_{surrounding}^4 \):**
- Calculate \( T^4 \):
\[ T^4 = (900)^4 = 6.561 \times 10^{8} \, \text{K}^4 \]
- Calculate \( T_{surrounding}^4 \):
\[ T_{surrounding}^4 = (300)^4 = 8.1 \times 10^{6} \, \text{K}^4 \]

4. **Substitute values into the power formula:**
Now we can substitute \( A \), \( T^4 \), and \( T_{surrounding}^4 \) into the power formula:
\[ P = \sigma A (T^4 - T_{surrounding}^4) \]
\[ P = 5.68 \times 10^{-8} \times (2 \pi \times 10^{-3}) \times (6.561 \times 10^{8} - 8.1 \times 10^{6}) \]

5. **Calculate the difference \( T^4 - T_{surrounding}^4 \):**
\[ T^4 - T_{surrounding}^4 = 6.561 \times 10^{8} - 8.1 \times 10^{6} \approx 6.4809 \times 10^{8} \]

6. **Calculate the power \( P \):**
\[ P = 5.68 \times 10^{-8} \times (2 \pi \times 10^{-3}) \times (6.4809 \times 10^{8}) \]
\[ P \approx 5.68 \times 10^{-8} \times 6.2832 \times 10^{-3} \times 6.4809 \times 10^{8} \]
\[ P \approx 230 \, \text{W} \]

### Final Answer:
The heat radiated per second by the wire is approximately **230 watts**.